3.89.85 \(\int \frac {1}{25} (25+(900+1050 x+450 x^2+100 x^3+(60+60 x+30 x^2) \log (2)+2 x \log ^2(2)) \log ^2(9)) \, dx\)

Optimal. Leaf size=28 \[ x+x^2 \left (x+\frac {3 (2+x)}{x}+\frac {\log (2)}{5}\right )^2 \log ^2(9) \]

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Rubi [B]  time = 0.04, antiderivative size = 80, normalized size of antiderivative = 2.86, number of steps used = 5, number of rules used = 2, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {12, 6} \begin {gather*} x^4 \log ^2(9)+\frac {2}{5} x^3 \log (2) \log ^2(9)+6 x^3 \log ^2(9)+\frac {1}{25} x^2 \left (525+\log ^2(2)\right ) \log ^2(9)+\frac {6}{5} x^2 \log (2) \log ^2(9)+x+\frac {12}{5} x \log (2) \log ^2(9)+36 x \log ^2(9) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25 + (900 + 1050*x + 450*x^2 + 100*x^3 + (60 + 60*x + 30*x^2)*Log[2] + 2*x*Log[2]^2)*Log[9]^2)/25,x]

[Out]

x + 36*x*Log[9]^2 + 6*x^3*Log[9]^2 + x^4*Log[9]^2 + (12*x*Log[2]*Log[9]^2)/5 + (6*x^2*Log[2]*Log[9]^2)/5 + (2*
x^3*Log[2]*Log[9]^2)/5 + (x^2*(525 + Log[2]^2)*Log[9]^2)/25

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{25} \int \left (25+\left (900+1050 x+450 x^2+100 x^3+\left (60+60 x+30 x^2\right ) \log (2)+2 x \log ^2(2)\right ) \log ^2(9)\right ) \, dx\\ &=x+\frac {1}{25} \log ^2(9) \int \left (900+1050 x+450 x^2+100 x^3+\left (60+60 x+30 x^2\right ) \log (2)+2 x \log ^2(2)\right ) \, dx\\ &=x+\frac {1}{25} \log ^2(9) \int \left (900+450 x^2+100 x^3+\left (60+60 x+30 x^2\right ) \log (2)+x \left (1050+2 \log ^2(2)\right )\right ) \, dx\\ &=x+36 x \log ^2(9)+6 x^3 \log ^2(9)+x^4 \log ^2(9)+\frac {1}{25} x^2 \left (525+\log ^2(2)\right ) \log ^2(9)+\frac {1}{25} \left (\log (2) \log ^2(9)\right ) \int \left (60+60 x+30 x^2\right ) \, dx\\ &=x+36 x \log ^2(9)+6 x^3 \log ^2(9)+x^4 \log ^2(9)+\frac {12}{5} x \log (2) \log ^2(9)+\frac {6}{5} x^2 \log (2) \log ^2(9)+\frac {2}{5} x^3 \log (2) \log ^2(9)+\frac {1}{25} x^2 \left (525+\log ^2(2)\right ) \log ^2(9)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.03, size = 59, normalized size = 2.11 \begin {gather*} x+x^4 \log ^2(9)+\frac {12}{5} x (15+\log (2)) \log ^2(9)+\frac {2}{5} x^3 (15+\log (2)) \log ^2(9)+\frac {1}{25} x^2 \left (525+30 \log (2)+\log ^2(2)\right ) \log ^2(9) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25 + (900 + 1050*x + 450*x^2 + 100*x^3 + (60 + 60*x + 30*x^2)*Log[2] + 2*x*Log[2]^2)*Log[9]^2)/25,x
]

[Out]

x + x^4*Log[9]^2 + (12*x*(15 + Log[2])*Log[9]^2)/5 + (2*x^3*(15 + Log[2])*Log[9]^2)/5 + (x^2*(525 + 30*Log[2]
+ Log[2]^2)*Log[9]^2)/25

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fricas [A]  time = 0.53, size = 51, normalized size = 1.82 \begin {gather*} \frac {4}{25} \, {\left (25 \, x^{4} + x^{2} \log \relax (2)^{2} + 150 \, x^{3} + 525 \, x^{2} + 10 \, {\left (x^{3} + 3 \, x^{2} + 6 \, x\right )} \log \relax (2) + 900 \, x\right )} \log \relax (3)^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(4/25*(2*x*log(2)^2+(30*x^2+60*x+60)*log(2)+100*x^3+450*x^2+1050*x+900)*log(3)^2+1,x, algorithm="fric
as")

[Out]

4/25*(25*x^4 + x^2*log(2)^2 + 150*x^3 + 525*x^2 + 10*(x^3 + 3*x^2 + 6*x)*log(2) + 900*x)*log(3)^2 + x

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giac [A]  time = 0.19, size = 51, normalized size = 1.82 \begin {gather*} \frac {4}{25} \, {\left (25 \, x^{4} + x^{2} \log \relax (2)^{2} + 150 \, x^{3} + 525 \, x^{2} + 10 \, {\left (x^{3} + 3 \, x^{2} + 6 \, x\right )} \log \relax (2) + 900 \, x\right )} \log \relax (3)^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(4/25*(2*x*log(2)^2+(30*x^2+60*x+60)*log(2)+100*x^3+450*x^2+1050*x+900)*log(3)^2+1,x, algorithm="giac
")

[Out]

4/25*(25*x^4 + x^2*log(2)^2 + 150*x^3 + 525*x^2 + 10*(x^3 + 3*x^2 + 6*x)*log(2) + 900*x)*log(3)^2 + x

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maple [A]  time = 0.04, size = 53, normalized size = 1.89




method result size



default \(\frac {4 \ln \relax (3)^{2} \left (x^{2} \ln \relax (2)^{2}+\ln \relax (2) \left (10 x^{3}+30 x^{2}+60 x \right )+25 x^{4}+150 x^{3}+525 x^{2}+900 x \right )}{25}+x\) \(53\)
gosper \(\frac {x \left (4 \ln \relax (2)^{2} \ln \relax (3)^{2} x +40 \ln \relax (2) \ln \relax (3)^{2} x^{2}+100 x^{3} \ln \relax (3)^{2}+120 x \ln \relax (2) \ln \relax (3)^{2}+600 x^{2} \ln \relax (3)^{2}+240 \ln \relax (2) \ln \relax (3)^{2}+2100 x \ln \relax (3)^{2}+3600 \ln \relax (3)^{2}+25\right )}{25}\) \(76\)
norman \(\left (\frac {8 \ln \relax (2) \ln \relax (3)^{2}}{5}+24 \ln \relax (3)^{2}\right ) x^{3}+\left (\frac {48 \ln \relax (2) \ln \relax (3)^{2}}{5}+144 \ln \relax (3)^{2}+1\right ) x +\left (\frac {4 \ln \relax (3)^{2} \ln \relax (2)^{2}}{25}+\frac {24 \ln \relax (2) \ln \relax (3)^{2}}{5}+84 \ln \relax (3)^{2}\right ) x^{2}+4 x^{4} \ln \relax (3)^{2}\) \(77\)
risch \(\frac {4 x^{2} \ln \relax (3)^{2} \ln \relax (2)^{2}}{25}+\frac {8 \ln \relax (2) \ln \relax (3)^{2} x^{3}}{5}+4 x^{4} \ln \relax (3)^{2}+\frac {24 \ln \relax (2) \ln \relax (3)^{2} x^{2}}{5}+24 x^{3} \ln \relax (3)^{2}+\frac {48 x \ln \relax (2) \ln \relax (3)^{2}}{5}+84 x^{2} \ln \relax (3)^{2}+144 x \ln \relax (3)^{2}+x\) \(81\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(4/25*(2*x*ln(2)^2+(30*x^2+60*x+60)*ln(2)+100*x^3+450*x^2+1050*x+900)*ln(3)^2+1,x,method=_RETURNVERBOSE)

[Out]

4/25*ln(3)^2*(x^2*ln(2)^2+ln(2)*(10*x^3+30*x^2+60*x)+25*x^4+150*x^3+525*x^2+900*x)+x

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maxima [A]  time = 0.37, size = 51, normalized size = 1.82 \begin {gather*} \frac {4}{25} \, {\left (25 \, x^{4} + x^{2} \log \relax (2)^{2} + 150 \, x^{3} + 525 \, x^{2} + 10 \, {\left (x^{3} + 3 \, x^{2} + 6 \, x\right )} \log \relax (2) + 900 \, x\right )} \log \relax (3)^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(4/25*(2*x*log(2)^2+(30*x^2+60*x+60)*log(2)+100*x^3+450*x^2+1050*x+900)*log(3)^2+1,x, algorithm="maxi
ma")

[Out]

4/25*(25*x^4 + x^2*log(2)^2 + 150*x^3 + 525*x^2 + 10*(x^3 + 3*x^2 + 6*x)*log(2) + 900*x)*log(3)^2 + x

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mupad [B]  time = 0.07, size = 62, normalized size = 2.21 \begin {gather*} 4\,{\ln \relax (3)}^2\,x^4+\frac {4\,{\ln \relax (3)}^2\,\left (30\,\ln \relax (2)+450\right )\,x^3}{75}+\frac {2\,{\ln \relax (3)}^2\,\left (60\,\ln \relax (2)+2\,{\ln \relax (2)}^2+1050\right )\,x^2}{25}+\left (\frac {4\,{\ln \relax (3)}^2\,\left (60\,\ln \relax (2)+900\right )}{25}+1\right )\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*log(3)^2*(1050*x + log(2)*(60*x + 30*x^2 + 60) + 2*x*log(2)^2 + 450*x^2 + 100*x^3 + 900))/25 + 1,x)

[Out]

4*x^4*log(3)^2 + x*((4*log(3)^2*(60*log(2) + 900))/25 + 1) + (4*x^3*log(3)^2*(30*log(2) + 450))/75 + (2*x^2*lo
g(3)^2*(60*log(2) + 2*log(2)^2 + 1050))/25

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sympy [B]  time = 0.08, size = 88, normalized size = 3.14 \begin {gather*} 4 x^{4} \log {\relax (3 )}^{2} + x^{3} \left (\frac {8 \log {\relax (2 )} \log {\relax (3 )}^{2}}{5} + 24 \log {\relax (3 )}^{2}\right ) + x^{2} \left (\frac {4 \log {\relax (2 )}^{2} \log {\relax (3 )}^{2}}{25} + \frac {24 \log {\relax (2 )} \log {\relax (3 )}^{2}}{5} + 84 \log {\relax (3 )}^{2}\right ) + x \left (1 + \frac {48 \log {\relax (2 )} \log {\relax (3 )}^{2}}{5} + 144 \log {\relax (3 )}^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(4/25*(2*x*ln(2)**2+(30*x**2+60*x+60)*ln(2)+100*x**3+450*x**2+1050*x+900)*ln(3)**2+1,x)

[Out]

4*x**4*log(3)**2 + x**3*(8*log(2)*log(3)**2/5 + 24*log(3)**2) + x**2*(4*log(2)**2*log(3)**2/25 + 24*log(2)*log
(3)**2/5 + 84*log(3)**2) + x*(1 + 48*log(2)*log(3)**2/5 + 144*log(3)**2)

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